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STD 12 - 5. continuity and differentiation — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 5. continuity and differentiation4 Marks
MCQ
If $y^{2}+\log _{e}\left(\cos ^{2} x\right)=y, x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right),$ then
  • ✓
    $\operatorname{|y}^{\prime \prime}(0) \mid=2$
  • B
    $\left|y^{\prime}(0)\right|+\left|y^{\prime \prime}(0)\right|=3$
  • C
    $\operatorname{|y}^{\prime}(0)\left|+\operatorname{|y}^{\prime \prime}(0)\right|=1$
  • D
    $y^{\prime \prime}(0)=0$
✓

Answer

Correct option: A.
$\operatorname{|y}^{\prime \prime}(0) \mid=2$
a
$y ^{2}+\ln \left(\cos ^{2} x \right)= y \quad x \in\left(-\frac{\pi}{2}, \frac{\pi}{2}\right)$

for $x=0 \quad y=0$ or 1

Differentiating wrt x

$\Rightarrow 2 y y^{\prime}-2 \tan x=y^{\prime}$

$\text { At }(0,0) y^{\prime}=0$

At $(0,1) y^{\prime}=0$

Differentiating wrt $x$

$2 y y^{\prime \prime}+2\left(y^{\prime}\right)^{2}-2 \sec ^{2} x=y^{\prime \prime}$

At $(0,0) \quad y^{\prime \prime}=-2$

At $(0,1) \quad y^{\prime \prime}=2$

$\therefore \quad \operatorname{ly}^{\prime \prime}(0) \mid=2$

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